I have to show that $[0,\infty)^2$ is not a differentiable manifold. The problem is $(0,0)$ (because there doesn't exist a diffeomorphism between $[0,\infty)^2$ and $R^2$) but I don't know how to show that.
$[0,\infty)^2$ = $[0,\infty) \times [0,\infty)$ (the non-negative part of the axes in $R^2$)
And then I am searching for a homeomorphism between $[0,\infty)^2$ and $H^2 := \{x=(x_1,x_2) \in R : x_2 ≥ 0 \}$ that is a diffeomorphism restricted to $[0,\infty)^2 - \{(0,0)\}$.
Thanks muchly!
If it was a manifold, then you have a chart $f:U\subset [0,\infty]^2\rightarrow V\subset R^2$ which contains $(0,0)$. $U$ can be supposed of the form $[0,c)^2$. Thus $V$ is open and contractible since $f$ is an homeomorphism. But the fundamental group of $V-f((0,0)$ is not trivial (the fundamental group of an open subset $W$ of the plan without a point $u$ is not trivial. To see this, consider $B(u,r)\subset W$. You cannot deform the circle of center $u$ and radius $r/2$ which is contained in $W-\{u\}$ to a point). consider and $U-(0,0)$ its fundamental group is trivial.